# From derivative of $f$ to squaring $f$

I am dealing with a math problem which I can't resolve.

Given $f:\mathbb{R}\to\mathbb{R}$ twice differentiable with the following property:

$$2x f(x)\geq f'(0)-f'(x)$$

Show that

$$f''(0)+f^2(0)\geq -1$$

I've been trying to use the derivative definition:

$$\lim \limits_{x \to x_0} \frac{f(x) -f(x_0)}{x-x_0}$$

But I got no result.

• Please use $signs for inline math-formatting as well as for display; Use \geq for >=, and \leq for <=. Also, this is a math forum, and * is not used for multiplication. Aug 22, 2018 at 13:34 ## 3 Answers$f''(0)$exists because$f$is twice differentiable. By definition$f''(0)=\lim\limits_{x\rightarrow0}\frac{f'(x)-f'(0)}{x}$You can minorate the fraction using the given property$2xf(x)\geq f'(0)-f'(x)$: for$x>0$,$\frac{f'(x)-f'(0)}{x}\geq-2f(x)$The LHS has a limit in$0$because$f$is twice differentiable. Taking the limits of continuous functions preserves inequalities, hence:$f''(0)\geq-2f(0)$From there$f''(0)+f^2(0)\geq-2f(0)+f^2(0)=(1-f(0))^2-1$And you can conclude using that a square is always non-negative. • @Evargalo thank you very much! Aug 22, 2018 at 14:03 hint For$x>0,$$$-2f(x)\le \frac{f'(x)-f'(0)}{x}$$ thus $$f''(0^+)\ge - 2f(0)$$ for$x<0\$,

$$-2f(x)\ge \frac{f'(x)-f(0)}{x}$$

$$\implies f''(0^-)\le -2f(0)$$

so $$f''(0)=-2f(0)$$

finally $$-2f(0)+f^2(0)+1=(f(0)-1)^2\ge 0$$

Note that $$f'(x)-f'(0)\ge-2xf(x)$$ and hence $$\frac{f'(x)-f'(0)}{x}\ge-2f(x), \forall x>0$$ and $$\frac{f'(x)-f'(0)}{x}\le-2f(x), \forall x<0.$$ So $$f''_+(0)=\lim_{x\to0^+}\frac{f'(x)-f'(0)}{x}\ge-2f(0) \tag{1}$$ and $$f''_-(0)=\lim_{x\to0^-}\frac{f'(x)-f'(0)}{x}\le-2f(0). \tag{2}$$ From (1)(2), one has $$f''(0)=-2f(0)$$ and hence $$f''(0)+f^2(0)=-2f(0)+f^2(0)=(f(0)-1))^2-1\ge-1.$$